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I try to identify $$S(z):= \sum_{n=1}^\infty \frac{1}{n} (z-2)^n$$ By the ratio test, $S(z)$ converges when $|z-2|<1$.

This looks like a Taylor series of $\ln()$, so upon checking, I found that, if we were just in the real case, the sum would evaluate to $-\ln(3-z)$.

However, I'm doing complex analysis right now, and it doesn't seem to make sense to me that it would be evaluated to be $\ln$, since the only logarithm I've met in complex analysis is $$\mathrm{Ln}(z):= \ln|z| + i\mathrm{Arg}(z)$$

What would the sum $S(z)$ be then?

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    $\begingroup$ $$|z-2|<1\implies S(z)=-\ln|3-z|-i\mathrm{Arg}(3-z)$$ $\endgroup$ – Did Aug 6 '17 at 13:34
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Appropriately, we have

$$S(z+2)=\sum_{n=1}^\infty\frac{z^n}n=-\operatorname{Log}(1-z)$$

Indeed, note for $|z-2|<1$ and $z+2=x+iy$, we have

$$\arg(x+iy)=\operatorname{atan2}(y,x)\in(-\pi/2,\pi/2)$$

which is not an entire branch. That is, we have

$$S(z+2)=-\ln|1-z|-i\arg(z)$$

For any choice of $\arg(z)\in(-\pi/2,\pi/2)$. The principle branch works as well.

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